If $y = x(x - 3)^2$ decreases for the values of $x$ given by

$1 < x < 3$

The correct answer is Option (1) → $1 < x < 3$ ##

We have,

$y = x(x - 3)^2$

$∴\frac{dy}{dx} = x \cdot 2(x - 3) \cdot 1 + (x - 3)^2 \cdot 1$

$= 2x^2 - 6x + x^2 + 9 - 6x = 3x^2 - 12x + 9$

$= 3(x^2 - 3x - x + 3) = 3(x - 3)(x - 1)$

So, $y = x(x - 3)^2$ decreases for $(1, 3)$.

$[\text{since, } y' < 0 \text{ for all } x \in (1, 3), \text{ hence } y \text{ is decreasing on } (1, 3)]$